Q: Which of the following equations has \$x = 2\$ as a solution?

\$\textbf{The value of the variable in an equation which satisfies the equation is called a solution to the}\$ \$\textbf{ equation.}\$Let us now try and check the value of \$x=2\$ for all the four given equationsFor option (A)\$\begin{aligned}&x+2=5 \\&2+2 \neq 5\end{aligned}\$\$\textbf{ Does not satisfy}\$For option (B)\$\begin{aligned}&x-2=0 \\&2-2=0\end{aligned}\$\$\textbf{ Satisfies.}\$For option (C)\$\begin{aligned}&2 x+1=0 \\&(2 \times 2)+1 \neq 0\end{aligned}\$\$\textbf{ Does not satisfy}\$For option (D)\$\begin{aligned}&x+3=6 \\&2+3 \neq 6\end{aligned}\$\$\textbf{ Does not satisfy}\$\$\textbf{Clearly option B is correct as it gives us the Left hand side equal to the Right hand side.}\$

Q: \$\textbf{State whether the statement is True or False.}\$\$5\$ is the solution of the equation \$3x+2=17\$.

Consider the given equation, \$3 x+2=17\$On transposing \$2\$ to RHS.We get,\$\Rightarrow 3 x=17-2\$\$\Rightarrow 3 x=15\$\$\Rightarrow x=\frac{15}{3}\$\$\Rightarrow x=5\$\$\therefore 5 \textbf{ is the solution of the equation } 3 x+2=17\$.\$\textbf{Hence, the given statement is true.}\$

Q: \$x = -1\$ is a solution of the equation

\$x = -1\$\$2(x-2) + 6 =0\$\$\implies\, 2(-1 -2) + 6 =0\$\$\implies\, 2 \times (-3) + 6= 0\$\$\implies\, -6 + 6 =0\$

Q: \$x = 5, y = 2\$ is a solution of the linear equation

\$\textbf{Step 1: Check which of the equation passes through (5,2).}\$ \$\text{The solution should satisfy the appropriate equation.}\$ \$\text{So, let us investigate each option by substituting }x=5\$ & \$y=2\$ \$(A)\$ \$x+2y=7\$. \$\text{L.H.S. }=5+2\times 2=9\neq\$ \$\text{R.H.S.}\$ \$ \therefore\$ \$\text{It is not the appropriate equation.}\$ \$(B)\$ \$5x+2y=7\$ \$\text{L.H.S. }=5\times 5+2\times 2=29\neq\$ \$\text{R.H.S.}\$ \$\therefore \$ \$\text{It is not the appropriate equation.}\$ \$(C)\$ \$x+y=7\$ \$\text{L.H.S. }=5+2=7=\$ \$\text{R.H.S}\$ \$\therefore\$ \$\text{It is the appropriate equation.}\$ \$(D)\$ \$ 5x+y=7\$ \$\text{L.H.S. }=5\times 5+2=27\neq\$ \$\text{R.H.S.}\$ \$\therefore \$ \$\text{It is not the appropriate equation.}\$\$\textbf{Hence, option C is correct.}\$

Q: Which equation among the following don't have \$(-1)\$ as a solution?

\${\textbf{Step -1: Considering option A.}}\$ \${\text{We have given,}}\$ \$x +1 = 0\$ \${\text{Putting }}x = - 1{\text{ we get,}}\$ \$ - 1 +1 = 0\$ \$ \Rightarrow 0 = 0\$ \${\text{Which is possible}}{\text{.}}\$ \${\text{Here, L}}{\text{.H}}{\text{.S}} = {\text{R}}{\text{.H}}{\text{.S }}\$ \${\text{So, - 1 is a solution of }}x +1 = 0.\$\${\textbf{Step -2: Considering option B.}}\$ \${\text{We have given,}}\$ \$3x + 4 = 1\$ \${\text{Putting }}x = - 1{\text{ we get,}}\$ \$3\left( { - 1} \right) + 4 = 1\$ \$ \Rightarrow - 3 + 4 = 1\$ \$ \Rightarrow 1 = 1\$ \${\text{Which is possible}}{\text{.}}\$ \${\text{Here, L}}{\text{.H}}{\text{.S }} = {\text{ R}}{\text{.H}}{\text{.S }}\$ \${\text{So, - 1 is a solution of 3}}x + 4 = 1.\$\${\textbf{Step -3: Considering option C.}}\$ \${\text{We have given,}}\$ \$5x + 7 = 2\$ \${\text{Putting }}x = - 1{\text{ we get,}}\$ \$5\left( { - 1 } \right) + 7 = 2\$ \$ \Rightarrow -5+ 7 = 2\$ \$ \Rightarrow 2 = 2\$ \${\text{Which is possible}}{\text{.}}\$ \${\text{Here, L}}{\text{.H}}{\text{.S = R}}{\text{.H}}{\text{.S }}\$ \${\text{So, - 1 is a solution of }}5x+7 =2.\$\${\textbf{Step -4: Considering option D.}}\$ \${\text{We have given,}}\$ \$x -1 = 2\$ \${\text{Putting }}x = - 1{\text{ we get,}}\$ \$-1-1= 2\$ \$ \Rightarrow - 2 = 2\$ \${\text{Which is not possible}}{\text{.}}\$ \${\text{Here, L}}{\text{.H}}{\text{.S}} \ne {\text{R}}{\text{.H}}{\text{.S }}\$ \${\text{So, - 1 is not a solution of }}x -1 = 2.\$\${\textbf{ Hence, option D. }}\mathbf{ x-1 =2}{\textbf{ is correct answer.}}\$

Q: Verify that \$-(-x)=x\$ for (i) \$x=\dfrac {11}{15}\$ (ii) \$x=-\dfrac {13}{17}\$

i) L.H.S \$=-(-x)\$\$=-\left (-\dfrac {11}{15}\right) = \dfrac {11}{15}\$ R.H.S \$= x = \dfrac {11}{15}\$ \$\therefore\$ L.H.S = R.H.S Hence, verified.i) L.H.S \$=-(-x)\$\$=-\left [-\dfrac {(-13)}{17}\right] = -\dfrac {13}{17}\$ R.H.S \$= x = -\dfrac {13}{17}\$ \$\therefore\$ L.H.S = R.H.S Hence, verified.

Q: The solution of the equation \$3(x - 4)= 30\$ is \$x =6\$

If \$x=6\$ is the solution then, \$3(x-4)=30\$ is satisfied by \$x=6\$LHS \$=3(6-4)=6\$RHS \$=30\$LHS \$\neq\$ RHS\$\implies x=6\$ is not the solution

Q: Verify that \$x=2\$ is a solution of the linear equation \$2x+7=13-x.\$

Solution:\$\textbf{Step 1: Find the value of L.H.S at }\$\$\boldsymbol{x=2}\$\$\quad\quad\quad\$Substitute \$x=2\$ in the left side of the given equation,\$\quad\quad\quad\$\$\begin{aligned}{}LHS& = 2x + 7\\& = 2(2) + 7\\ &= 11\end{aligned}\$\$\textbf{Step 2: Find the value of R.S.H at}\$\$\boldsymbol{x=2}\$\$\quad\quad\quad\$Substitute \$x=2\$ in the right side of the given equation,\$\quad\quad\quad\$\$\begin{aligned}{}RHS &= 13 - 2\\ &= 11\end{aligned}\$\$\quad\quad\quad\$Since, \$LHS = RHS=11\$ \$\ \$ at \$x=2\$.\$\textbf{Hence, \$2\$ is the solution of the given linear equation.}\$

Q: Check whether \$p=0\$ is a solution to \$4p-3 = 13\$ or not.

Given: \$4p-3= 13\$Putting \$p=0\$ in L.H.S.,\$\Rightarrow 4(0)-3=-3\$\$\because L.H.S. \neq R.H.S.\$ \$\therefore p= 0\$ is not the solution of given equation.

Question 1

Which of the following equations has $x = 2$ as a solution?

Accepted Answer

$\textbf{The value of the variable in an equation which satisfies the equation is called a solution to the}$ $\textbf{ equation.}$Let us now try and check the value of $x=2$ for all the four given equationsFor option (A)$\begin{aligned}&x+2=5 \&2+2 \neq 5\end{aligned}$$\textbf{ Does not satisfy}$For option (B)$\begin{aligned}&x-2=0 \&2-2=0\end{aligned}$$\textbf{ Satisfies.}$For option (C)$\begin{aligned}&2 x+1=0 \&(2 \times 2)+1 \neq 0\end{aligned}$$\textbf{ Does not satisfy}$For option (D)$\begin{aligned}&x+3=6 \&2+3 \neq 6\end{aligned}$$\textbf{ Does not satisfy}$$\textbf{Clearly option B is correct as it gives us the Left hand side equal to the Right hand side.}$

Question 2

$\textbf{State whether the statement is True or False.}$$5$ is the solution of the equation $3x+2=17$.

Accepted Answer

Consider the given equation, $3 x+2=17$On transposing $2$ to RHS.We get,$\Rightarrow 3 x=17-2$$\Rightarrow 3 x=15$$\Rightarrow x=\frac{15}{3}$$\Rightarrow x=5$$\therefore 5 \textbf{ is the solution of the equation } 3 x+2=17$.$\textbf{Hence, the given statement is true.}$

Question 3

$x = -1$ is a solution of the equation

Accepted Answer

$x = -1$$2(x-2) + 6 =0$$\implies\,  2(-1 -2) + 6 =0$$\implies\,  2 \times (-3) + 6= 0$$\implies\,  -6 + 6 =0$

Question 4

$x = 5, y = 2$ is a solution of the linear equation

Accepted Answer

$\textbf{Step 1: Check which of the equation passes through (5,2).}$                   $\text{The solution should satisfy the appropriate equation.}$                 $\text{So, let us investigate each option by substituting }x=5$ & $y=2$                 $(A)$      $x+2y=7$.                 $\text{L.H.S. }=5+2\times 2=9\neq$ $\text{R.H.S.}$                 $ \therefore$ $\text{It is not the appropriate equation.}$                 $(B)$      $5x+2y=7$                 $\text{L.H.S. }=5\times 5+2\times 2=29\neq$ $\text{R.H.S.}$                 $\therefore $ $\text{It is not the appropriate equation.}$                 $(C)$     $x+y=7$                  $\text{L.H.S. }=5+2=7=$ $\text{R.H.S}$                 $\therefore$ $\text{It  is the appropriate equation.}$                 $(D)$     $ 5x+y=7$                  $\text{L.H.S. }=5\times 5+2=27\neq$ $\text{R.H.S.}$                 $\therefore $ $\text{It is not the appropriate equation.}$$\textbf{Hence, option C is correct.}$

Question 5

Which equation among the following don't have $(-1)$ as a solution?

Accepted Answer

${\textbf{Step -1: Considering option A.}}$               ${\text{We have given,}}$ $x +1 = 0$               ${\text{Putting }}x =  - 1{\text{ we get,}}$               $ - 1 +1 = 0$               $ \Rightarrow  0 = 0$ ${\text{Which is possible}}{\text{.}}$               ${\text{Here, L}}{\text{.H}}{\text{.S}} = {\text{R}}{\text{.H}}{\text{.S }}$               ${\text{So,  - 1 is a solution of }}x +1 = 0.$${\textbf{Step -2: Considering option B.}}$               ${\text{We have given,}}$ $3x + 4 = 1$               ${\text{Putting }}x =  - 1{\text{ we get,}}$               $3\left( { - 1} \right) + 4 = 1$               $ \Rightarrow  - 3 + 4 = 1$               $ \Rightarrow 1 = 1$ ${\text{Which is possible}}{\text{.}}$               ${\text{Here, L}}{\text{.H}}{\text{.S }} = {\text{ R}}{\text{.H}}{\text{.S }}$               ${\text{So,  - 1 is a solution of 3}}x + 4 = 1.$${\textbf{Step -3: Considering option C.}}$               ${\text{We have given,}}$ $5x + 7 = 2$               ${\text{Putting }}x =  - 1{\text{ we get,}}$               $5\left( { - 1 } \right) + 7 = 2$               $ \Rightarrow -5+ 7 = 2$               $ \Rightarrow  2 = 2$ ${\text{Which is possible}}{\text{.}}$               ${\text{Here, L}}{\text{.H}}{\text{.S = R}}{\text{.H}}{\text{.S }}$               ${\text{So,  - 1 is a solution of }}5x+7 =2.$${\textbf{Step -4: Considering option D.}}$               ${\text{We have given,}}$ $x -1 = 2$               ${\text{Putting }}x =  - 1{\text{ we get,}}$               $-1-1= 2$               $ \Rightarrow  - 2 = 2$ ${\text{Which is not possible}}{\text{.}}$               ${\text{Here, L}}{\text{.H}}{\text{.S}} \ne {\text{R}}{\text{.H}}{\text{.S }}$               ${\text{So,  - 1 is not a solution of }}x -1 = 2.$${\textbf{ Hence, option D. }}\mathbf{ x-1 =2}{\textbf{ is correct answer.}}$

Question 6

Verify that $-(-x)=x$ for (i) $x=\dfrac {11}{15}$         (ii) $x=-\dfrac {13}{17}$

Accepted Answer

i) L.H.S $=-(-x)$$=-\left (-\dfrac {11}{15}\right) = \dfrac {11}{15}$   R.H.S $= x = \dfrac {11}{15}$   $\therefore$ L.H.S = R.H.S    Hence, verified.i) L.H.S $=-(-x)$$=-\left [-\dfrac {(-13)}{17}\right] = -\dfrac {13}{17}$   R.H.S $= x = -\dfrac {13}{17}$   $\therefore$ L.H.S = R.H.S    Hence, verified.

Question 7

The solution of the equation $3(x - 4)= 30$ is $x =6$

Accepted Answer

If $x=6$ is the solution then, $3(x-4)=30$ is satisfied by $x=6$LHS $=3(6-4)=6$RHS $=30$LHS $\neq$ RHS$\implies x=6$ is not the solution

Question 8

Verify that $x=2$ is a solution of the linear equation $2x+7=13-x.$

Accepted Answer

Solution:$\textbf{Step 1: Find the value of L.H.S at }$$\boldsymbol{x=2}$$\quad\quad\quad$Substitute  $x=2$ in the left side of the given equation,$\quad\quad\quad$$\begin{aligned}{}LHS& = 2x + 7\& = 2(2) + 7\ &= 11\end{aligned}$$\textbf{Step 2: Find the value of R.S.H at}$$\boldsymbol{x=2}$$\quad\quad\quad$Substitute $x=2$ in the right side of the given equation,$\quad\quad\quad$$\begin{aligned}{}RHS &= 13 - 2\ &= 11\end{aligned}$$\quad\quad\quad$Since, $LHS = RHS=11$ $\ $ at $x=2$.$\textbf{Hence, $2$ is the solution of the given linear equation.}$

Question 9

Check whether $p=0$ is a solution to $4p-3 = 13$ or not.

Accepted Answer

Given: $4p-3= 13$Putting $p=0$ in L.H.S.,$\Rightarrow 4(0)-3=-3$$\because L.H.S. \neq R.H.S.$ $\therefore p= 0$ is not the solution of given equation.

Question 10

Use trial and error method, find the value of $x$ if $5x = 25$.

Accepted Answer

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Quick Summary With Stories

Checking for Solutions of Equation

Which of the following equations has $x = 2$ as a solution?

$State whether the statement is True or False.$
$5$ is the solution of the equation $3 x + 2 = 17$ .

$x = - 1$ is a solution of the equation

$x = 5, y = 2$ is a solution of the linear equation

Which equation among the following don't have $(- 1)$ as a solution?

Verify that $- (- x) = x$ for
(i) $x = \frac{1 1}{1 5}$ (ii) $x = - \frac{1 3}{1 7}$

The solution of the equation $3 (x - 4) = 30$ is $x = 6$

Verify that $x = 2$ is a solution of the linear equation $2 x + 7 = 13 - x .$

Check whether $p = 0$ is a solution to $4 p - 3 = 13$ or not.

Use trial and error method, find the value of $x$ if $5 x = 25$ .

Learn with Videos

Quick Summary With Stories

Checking for Solutions of Equation

Which of the following equations has x=2 as a solution?

State whether the statement is True or False. 5 is the solution of the equation 3x+2=17.

x=−1 is a solution of the equation

x=5,y=2 is a solution of the linear equation

Which equation among the following don't have (−1) as a solution?

Verify that −(−x)=x for (i) x=1511​ (ii) x=−1713​

The solution of the equation 3(x−4)=30 is x=6

Verify that x=2 is a solution of the linear equation 2x+7=13−x.

Check whether p=0 is a solution to 4p−3=13 or not.

Use trial and error method, find the value of x if 5x=25.

Which of the following equations has $x = 2$ as a solution?

$State whether the statement is True or False.$
$5$ is the solution of the equation $3 x + 2 = 17$ .

$x = - 1$ is a solution of the equation

$x = 5, y = 2$ is a solution of the linear equation

Which equation among the following don't have $(- 1)$ as a solution?

Verify that $- (- x) = x$ for
(i) $x = \frac{1 1}{1 5}$ (ii) $x = - \frac{1 3}{1 7}$

The solution of the equation $3 (x - 4) = 30$ is $x = 6$

Verify that $x = 2$ is a solution of the linear equation $2 x + 7 = 13 - x .$

Check whether $p = 0$ is a solution to $4 p - 3 = 13$ or not.

Use trial and error method, find the value of $x$ if $5 x = 25$ .